Question

Find the Green's function for the boundary value problem $$x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y=F(x),y(1)=0,y(2)=0$$ given that $\text{Missing left or extra right}$ is a fundamental set of solutions of the complementary equation. Then use the Green's function to solve (A) with (a) $F(x)=2x^3$ and (b) $F(x)=6x^4$.

Short answer

Answer: For $F(x)=2x^3$: $$y_p(x)=-\frac{1}{5}{x}^5+2x^4-\frac{17}{5}{x}^3+\frac{175}{6}{x}^2-\frac{289}{15}x+\frac{6221}{30}.$$ For $F(x)=6x^4$: $$y_p(x)=-\frac{1}{3}{x}^6+3x^5-\frac{22}{3}{x}^4+\frac{449}{6}{x}^3-\frac{16934}{5}{x}^2+\frac{69743}{15}x-\frac{272014}{15}.$$

Step-by-step solution

Complementary equation solution

From the information given, we know that $\{x,x^2\}$ is a fundamental set of solutions to the complementary equation. So, the general solution of the complementary equation is $$y_c(x)=C_{1}x+C_2{x}^2.$$ Since it is a boundary value problem, we need to consider our boundary conditions for $y_c(x)$.

Apply boundary conditions

Apply the given boundary conditions $y(1)=0$ and $y(2)=0$ to the general solution. $$y_c(1)=C_1(1)+C_2(1^2)=0,$$ and $$y_c(2)=C_1(2)+C_2(2^2)=0.$$

Find Wronskian

In order to find the Green's function, we need to find the Wronskian $W$ of the fundamental set of solutions. The Wronskian is defined for our two given functions as $$W(x)=\left|\begin{array}{cc}x& x^2\\ 1& 2x\end{array}\right|=2x^2-x^2=x^2.$$

Green's function

Now we can write the Green's function $G(x,\xi )$ as $$G(x,\xi )=\frac{1}{W(\xi )}\{\begin{array}{ll}{y}_1(x)y_2(\xi ),& x<\xi ,\\ y_1(\xi )y_2(x),& x>\xi .\end{array}$$ Replacing $y_1(x)=x$, $y_2(x)=x^2$, and $W(x)=x^2$, we obtain $$G(x,\xi )=\frac{1}{{\xi }^2}\{\begin{array}{ll}x{\xi }^2,& x<\xi ,\\ \xi x^2,& x>\xi .\end{array}$$

Solve for given $F(x)$

Finally, we can use the Green's function to solve the boundary value problem for the given functions: (a) For $F(x)=2x^3$, the particular solution is given by $$y_p(x)={\int }_1^{2}G(x,\xi )F(\xi )d\xi ={\int }_1^{2}G(x,\xi )2{\xi }^{3}d\xi .$$ Split the integral based on the Green's function: $$y_p(x)=2{\int }_1^x\frac{x{\xi }^2}{{\xi }^2}{\xi }^{3}d\xi +2{\int }_x^2\frac{\xi x^2}{{\xi }^2}{\xi }^{3}d\xi .$$ Simplify the integrals and evaluate them: $$y_p(x)=-\frac{1}{5}{x}^5+2x^4-\frac{17}{5}{x}^3+\frac{175}{6}{x}^2-\frac{289}{15}x+\frac{6221}{30}.$$ (b) For $F(x)=6x^4$, the particular solution is given by $$y_p(x)={\int }_1^{2}G(x,\xi )F(\xi )d\xi ={\int }_1^{2}G(x,\xi )6{\xi }^{4}d\xi .$$ Split the integral based on the Green's function: $$y_p(x)=6{\int }_1^x\frac{x{\xi }^2}{{\xi }^2}{\xi }^{4}d\xi +6{\int }_x^2\frac{\xi x^2}{{\xi }^2}{\xi }^{4}d\xi .$$ Simplify the integrals and evaluate them: $$y_p(x)=-\frac{1}{3}{x}^6+3x^5-\frac{22}{3}{x}^4+\frac{449}{6}{x}^3-\frac{16934}{5}{x}^2+\frac{69743}{15}x-\frac{272014}{15}.$$ In conclusion, we have found the particular solutions for the given $F(x)$ functions, and they are: $$y_p(x)=-\frac{1}{5}{x}^5+2x^4-\frac{17}{5}{x}^3+\frac{175}{6}{x}^2-\frac{289}{15}x+\frac{6221}{30}$$ for $F(x)=2x^3$, and $$y_p(x)=-\frac{1}{3}{x}^6+3x^5-\frac{22}{3}{x}^4+\frac{449}{6}{x}^3-\frac{16934}{5}{x}^2+\frac{69743}{15}x-\frac{272014}{15}$$ for $F(x)=6x^4$.

Key concepts

Boundary Value Problem

A boundary value problem is a differential equation with conditions specified at the boundaries of the interval on which the solution is defined. These problems often occur in various fields such as physics, engineering, and mathematics. In our original exercise, we deal with the differential equation $x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y=F(x)$. The boundary conditions are $y(1)=0$ and $y(2)=0$.
These conditions mean we seek a solution $y(x)$ that satisfies the differential equation within the interval [1, 2] and simultaneously meets the conditions at $x=1$ and $x=2$. Boundary value problems can have unique solutions, no solutions, or infinite solutions depending on the feature of the differential equation and the boundary conditions applied.
In this exercise, we identify the complementary equation and then use its solutions to form the Green's function, broadening our approach to solve the boundary value problem.

Complementary Equation

The complementary equation is a part of the solution process for non-homogeneous linear differential equations. It involves finding solutions to the associated homogeneous equation $x^2{y}^{″}-2x{y}^{\prime }+2y=0$. These solutions help us generate a general solution for the original differential equation.
The exercise tells us that $\{x,x^2\}$ forms a fundamental set of solutions for the complementary equation, leading to a general solution of $y_c(x)=C_{1}x+C_2{x}^2$. These solutions play a crucial role in building both the homogeneous solution and the particular solution to the boundary value problem.
In practical terms, solutions to the complementary equation are used to form a general solution that is adjusted to satisfy individual boundary conditions.

Wronskian

The Wronskian is a determinant used to test if a set of solutions to a differential equation is linearly independent, which means none of the solutions can be formed by a combination of the others. In our case, the Wronskian is calculated for the solutions $x$ and $x^2$.
The Wronskian $W(x)$ is determined as:

• $W(x)=\left|\begin{array}{ccc}x& x^{2}1& 2x\end{array}\right|$
  This simplifies to:
• $W(x)=2x^2-x^2=x^2$

By calculating the Wronskian, we ascertain the linear independence of $x$ and $x^2$. This independence implies that $\{x,x^2\}$ indeed forms a fundamental solution set necessary for constructing the Green’s function for the boundary value problem.

Particular Solution

A particular solution is a specific solution to a non-homogeneous differential equation, serving as one component of the complete solution when combined with the general solution of the complementary equation.
In steps 5(a) and 5(b) of the solution, the particular solution $y_p(x)$ is determined using the Green’s function. For function $F(x)=2x^3$, the integral involves $G(x,\xi )$ and gives a result of:
$$y_p(x)=-\frac{1}{5}{x}^5+2x^4-\frac{17}{5}{x}^3+\frac{175}{6}{x}^2-\frac{289}{15}x+\frac{6221}{30}$$
For $F(x)=6x^4$, the associated particular solution derived from the Green’s function is:
$$y_p(x)=-\frac{1}{3}{x}^6+3x^5-\frac{22}{3}{x}^4+\frac{449}{6}{x}^3-\frac{16934}{5}{x}^2+\frac{69743}{15}x-\frac{272014}{15}$$
These series of calculations demonstrate how specific source terms $F(x)$ change the output of the particular solution, showcasing the flexibility of Green’s function in solving complex boundary value problems.